Frictional Energy: Solving for Air Resistance

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A 145 g baseball dropped from 8.0 m hits the ground at 7.80 m/s, prompting a discussion on calculating the average force of air resistance. The initial attempt to find air resistance using energy equations resulted in an incorrect value of 5.56 N. Suggestions include using kinematic equations to determine the actual acceleration of the ball and adjusting for air resistance. The approach involves calculating net acceleration and then multiplying by the mass to find the force of air resistance. Accurate arithmetic is essential for arriving at the correct answer.
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Frictional Energy??

Homework Statement



A 145 g baseball is dropped from a tree 8.0 m above the ground.
(b) If it actually hits the ground with a speed of 7.80 m/s, what is the magnitude of the average force of air resistance exerted on it?

Homework Equations



I've tried this problem several times, but I do not know how to find air resistance using the energy of the problem. Any help would be appreciated!

(air resistance) h= (delta)KE +(delta)PE



The Attempt at a Solution



(air resistance)=(m(.5(v^2)-gh))/h
Using this I got 5.56 N as an answer, but apparently that's wrong.

Any help whatsoever is appreciated!
 
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You can use the kinematic equations to find the actual acceleration of the ball. You know that minus any air resistance it should be the acceleration due to gravity g. You can then take g away from the acceleration you find to find the net acceleration due to air resistance. Multiply this by the mass of the ball to find the force.
 
gloryrentgirl said:
(air resistance)=(m(.5(v^2)-gh))/h
Using this I got 5.56 N as an answer, but apparently that's wrong.
Your method is OK. Check your arithmetic.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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